nice-calculus-0-1-dx-x-2-x-2-1-x-3-1-5-

Question Number 132741 by mnjuly1970 last updated on 16/Feb/21

\dots n i c e c a l c u l u s \dots

\int_{0^{}}^{1} \frac{d x}{(x - 2) {(x^{2} {(1 - x)}^{3})}^{\frac{1}{5}}} = ?

Commented by MJS_new last updated on 16/Feb/21

I get - 2^{1 / 10} π \sqrt{1 - \frac{\sqrt{5}}{5}} \approx - 2.50340750

using the substitution t = {(\frac{2 (1 - x)}{x})}^{1 / 5}

Commented by MJS_new last updated on 16/Feb/21

yes of course . forgot the “ -^{″}

Commented by Dwaipayan Shikari last updated on 16/Feb/21

Y e s s i r! B u t i t h i n k t h e r e s u l t s h o u l d b e - v e

Commented by mnjuly1970 last updated on 16/Feb/21

b r a v o m r M j

t h a n k y o u s o m u c h \dots

Answered by Dwaipayan Shikari last updated on 16/Feb/21

- \int_{0}^{1} t^{- \frac{3}{5}} {(1 - t)}^{- \frac{2}{5}} {(1 + t)}^{- 1} d t 1 - x = t

_{2} F_{1} (a, b; c; z) = \frac{Γ (c)}{Γ (c - b) Γ (b)} \int_{0}^{1} x^{b - 1} {(1 - x)}^{c - b - 1} {(1 - z x)}^{- a} d x

b - 1 = - \frac{3}{5} \Rightarrow b = \frac{2}{5}

c - b - 1 = \frac{- 2}{5} \Rightarrow c = 1

a = 1 z = - 1

- Γ (\frac{3}{5}) Γ (\frac{2}{5})_{2} F_{1} (1, \frac{2}{5}, 1, - 1) = - \int_{0}^{1} t^{- \frac{3}{5}} {(1 - t)}^{- \frac{2}{5}} {(1 + t)}^{- 1} d t

= - \frac{2 \sqrt{2} π}{\sqrt{5 - \sqrt{5}}}_{2} F_{1} (1, \frac{2}{5}, 1; - 1)

Commented by mnjuly1970 last updated on 16/Feb/21

m e r c e y

g r a t e f u l \dots

h y p e r g e o m e t r e y f u n c t i o n \dots